Q&A for How to Algebraically Find the Intersection of Two Lines

If that happens, you'll end up with a contradiction (like 1 = 2), which means that those two lines will never intersect.

I suspect that you copied this problem down wrong. I'll deal with what you wrote first, and then I'll talk about what I think you may have meant. As written, the first function says F(x)=2^2=12x+10. In other words, this is a simple one variable equation that simplifies to 4=12x+10. Then subtract 10 from both sides to get -6=12x. Finally, divide both sides by 12 to get -1/2 = x. You now have two different functions, each with a single variable. F(x): x=-1/2, and G(x): x=38. Any function that has only a single variable like this, x=__, is going to be a vertical straight line at that value. As a result, these two lines will never intersect, and there is no single solution for F(x) and G(x) simultaneously. That is not a very interesting solution, which makes me think you copied it wrong. I think that what you probably meant is F(x)=x^2 + 12x + 10. I think you wrote 2^2 instead of x^2, and then you changed a + symbol into an = symbol in the middle of the function. (The + and = are the same button on most keyboards.) This becomes a more interesting problem. You could now work on factoring the first function, but you don't need to do that much work. If you notice, the second function, G(x), is already solved. It is the single value, G(x)=38. This means that the graph of that function is a straight vertical line. At every point on the line, x=38. So to solve the system, just insert the value 38 for x in the first equation: F(x)=38^2+12(38)+10. This equals 1444+456+10, which is F(x)=1910. So the solution where those two graphs cross is x=38, y=1910. You can write the coordinate pair as (38,1910).

The lines could be x = 3 and y = 6.

If you are asking about two linear (straight line) equations, there will be only one point of intersection. This is explained in Method 1 above.

Because the x value of the specified points is unknown, you don't know where the specified points lie, and therefore you can't find either the y-intercept of the lines or their slope-intercept equations. Therefore, you cannot determine the point of intersection.

Remember that factoring only works with quadratic equations. If completing the square doesn't work, try using the quadratic equation and vice versa.

Isolate either variable yourself. For example, in the first equation, isolate and solve for x by subtracting 10y from both sides and then dividing both sides by 4. Isolate and solve for y by subtracting 4x from both sides and then dividing both sides by 10.

Because each equation represents a straight line, there will be just one point of intersection. In this case the easiest way to solve for x and y is to add the two equations together (by adding the left sides together, adding the right sides together, and setting the two sums equal to each other): (x+y) + (-x+y) = (-3) + (3). Then 2y = 0, and y = 0. Substitute the y value into either of the original equations to find the x value: x + 0 = -3, and x = -3. So the point of intersection of the two lines is (-3,0).

Use the quadratic equation -b(square root) b^2-4ac / 2a.

The intersection occurs at the point(s) where the two equations equal each other. So set one equation equal to the other, and solve for x. Then substitute that x value back into either equation to get the y value. You then have the x and y values of the point of intersection.

Y=Y so we can say that -5/6x - 9/6 = 4/3x - 8. We need to combine like terms and so you would add 9/6 to both sides ( -5/6x = 4/3x - 13/2 ). After that, we would subtract 4/3x from both sides ( -13/6x = -13/2 ). Because of the large fractions, let's multiply both sides by 6. ( -13x = -39 ). The X needs to be on it's own so we divide both sides by -13 ( x = 3 ). Plug in the X to either equation to solve for Y and you would get the answer (3,-4).

You can start by assuming that they do intersect, meaning P(x) = L(x) at some point. Then you can simplify so you have P(x) - L(x) = 0. On the left-hand side, you'll have a quadratic so you could try factoring, the quadratic formula, or graphing to see when it is equal to zero. If P(x) does not intersect L(x), then the roots of the left-hand side will be imaginary.

In this case, the easiest solution is by substitution. (See Solve Simultaneous Equations Using Substitution Method .) That yields an x value and a y value that represent the x- and y-coodinates of the point of intersection.

If by (4;0) you mean (4,0), (4,0) is a point, and points do not have points of intersection. (Only lines have points of intersection.)